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The maximum efficiency of a 100 kVA, signal phase transformer is 98% and occurs at 80% of full load at 0.8 P.F. If the leakage impedance of the transformer is 5%, find voltage regulation (in %) at rated load of 0.8 power factor lagging.
Correct answer is '3.8'. Can you explain this answer?
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Verified Answer
The maximum efficiency of a 100 kVA, signal phase transformer is 98% a...
Since maximum efficiency occurs at 80% of full load at 0.8 P.F
Output at ηmax = (100 × 0.8) × 0.8 = 64 kW
∴ Total loss = 65.3 – 64 = 1.3 kW
This loss is divided equally between cu and iron
∴ Cu loss at 80% of full load = 1.3/2 = 0.65 kW
Cu loss at full load = 0.65/82 = 1 kW
∴ % Regulation = [1 × 0.8 + 5 × 0.6] = 3.8%
Output at ηmax = (100 × 0.8) × 0.8 = 64 kW
∴ Total loss = 65.3 – 64 = 1.3 kW
This loss is divided equally between cu and iron
∴ Cu loss at 80% of full load = 1.3/2 = 0.65 kW
Cu loss at full load = 0.65/82 = 1 kW
∴ % Regulation = [1 × 0.8 + 5 × 0.6] = 3.8%
Most Upvoted Answer
The maximum efficiency of a 100 kVA, signal phase transformer is 98% a...
Given data:
- Power rating of transformer (S) = 100 kVA
- Maximum efficiency (η_max) = 98%
- Load at maximum efficiency (L_max) = 80% of full load
- Power factor at maximum efficiency (P.F_max) = 0.8
- Leakage impedance (Z_l) = 5%
To find: Voltage regulation (VR) at rated load of 0.8 power factor lagging
Formulae used:
1. Efficiency (η) = Output power / Input power
= Output power / (Output power + Losses)
= S × P.F × (1 - (Z_l / Z)) / (S × P.F)
2. Voltage regulation (VR) = (V_no-load - V_full-load) / V_full-load × 100%
= (I_2 × Z - V_full-load) / V_full-load × 100%
= (I_2 × Z / V_full-load - 1) × 100%
where I_2 = rated load current, Z = impedance of transformer
Calculations:
1. Efficiency:
Let's assume the voltage and current at maximum efficiency as V_max and I_max, respectively.
P_in = V_max × I_max × P.F_max
P_out = V_max × I_max
Losses = P_in - P_out
η_max = P_out / P_in = V_max × I_max / (V_max × I_max × 0.8 + Losses)
0.98 = V_max × I_max / (V_max × I_max × 0.8 + (V_max × I_max - V_max × I_max / η_max))
0.98 = V_max × I_max / (V_max × I_max × 0.8 + V_max × I_max - V_max × I_max / 0.98)
V_max / V_full-load = 1 - (1 - L_max) / η_max
V_max / V_full-load = 1 - (1 - 0.8) / 0.98
V_max / V_full-load = 0.8163
2. Voltage regulation:
I_2 = S / (V_full-load × P.F)
Z = V_full-load / (I_2 × 100 / 5) [since Z_l is given as a percentage of rated voltage]
VR = (I_2 × Z / V_full-load - 1) × 100%
VR = (S / (V_full-load × 0.8) × V_full-load / (I_2 × 100 / 5) / V_full-load - 1) × 100%
VR = (S / (V_full-load × 0.8) × 20 / I_2 - 1) × 100%
VR = (100 × 1000 / (V_full-load × 0.8) × 20 / (S / (V_full-load × 0.8))) - 1) × 100%
VR = (100 × 1000 / (V_full-load × 0.8) × 20 / 125 - 1) × 100% [substituting S = 100 kVA]
VR = 3.84% (approx.)
Therefore, the voltage regulation at rated
- Power rating of transformer (S) = 100 kVA
- Maximum efficiency (η_max) = 98%
- Load at maximum efficiency (L_max) = 80% of full load
- Power factor at maximum efficiency (P.F_max) = 0.8
- Leakage impedance (Z_l) = 5%
To find: Voltage regulation (VR) at rated load of 0.8 power factor lagging
Formulae used:
1. Efficiency (η) = Output power / Input power
= Output power / (Output power + Losses)
= S × P.F × (1 - (Z_l / Z)) / (S × P.F)
2. Voltage regulation (VR) = (V_no-load - V_full-load) / V_full-load × 100%
= (I_2 × Z - V_full-load) / V_full-load × 100%
= (I_2 × Z / V_full-load - 1) × 100%
where I_2 = rated load current, Z = impedance of transformer
Calculations:
1. Efficiency:
Let's assume the voltage and current at maximum efficiency as V_max and I_max, respectively.
P_in = V_max × I_max × P.F_max
P_out = V_max × I_max
Losses = P_in - P_out
η_max = P_out / P_in = V_max × I_max / (V_max × I_max × 0.8 + Losses)
0.98 = V_max × I_max / (V_max × I_max × 0.8 + (V_max × I_max - V_max × I_max / η_max))
0.98 = V_max × I_max / (V_max × I_max × 0.8 + V_max × I_max - V_max × I_max / 0.98)
V_max / V_full-load = 1 - (1 - L_max) / η_max
V_max / V_full-load = 1 - (1 - 0.8) / 0.98
V_max / V_full-load = 0.8163
2. Voltage regulation:
I_2 = S / (V_full-load × P.F)
Z = V_full-load / (I_2 × 100 / 5) [since Z_l is given as a percentage of rated voltage]
VR = (I_2 × Z / V_full-load - 1) × 100%
VR = (S / (V_full-load × 0.8) × V_full-load / (I_2 × 100 / 5) / V_full-load - 1) × 100%
VR = (S / (V_full-load × 0.8) × 20 / I_2 - 1) × 100%
VR = (100 × 1000 / (V_full-load × 0.8) × 20 / (S / (V_full-load × 0.8))) - 1) × 100%
VR = (100 × 1000 / (V_full-load × 0.8) × 20 / 125 - 1) × 100% [substituting S = 100 kVA]
VR = 3.84% (approx.)
Therefore, the voltage regulation at rated
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The maximum efficiency of a 100 kVA, signal phase transformer is 98% and occurs at 80% of full load at 0.8 P.F. If the leakage impedance of the transformer is 5%, find voltage regulation (in %) at rated load of 0.8 power factor lagging.Correct answer is '3.8'. Can you explain this answer?
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